如何限制表A中的行数，并有更多连接表B，表C，

Question

我有更多的桌子连接在一起：

module.exports = function (sequelize, DataTypes) {   class Document extends AT_ModelDb {       static associate(models) {       // define association here       models.Document.belongsTo       (models.DocumentType              ,{as: 'type'                     , foreignKey : 'type_id'            , targetKey  : "id"                       });       models.Document.belongsTo       (models.DocumentStatus            ,{as: 'status'                   , foreignKey : 'status_id'          , targetKey  : "id"                       });       models.Document.belongsTo       (models.Document                  ,{as: 'documentLinked'           , foreignKey : "documentLinked_id"  , targetKey  : 'id'                       });       models.Document.belongsTo       (models.DocumentBatch             ,{as: 'batch'                    , foreignKey : "batch_id"           , targetKey  : 'id'                       });       models.Document.belongsTo       (models.Project                   ,{as: 'project'                  , foreignKey : "project_id"         , targetKey  : 'id'                       });       models.Document.belongsTo       (models.Budget                    ,{as: 'budget'                   , foreignKey : 'budget_id'          , targetKey  : "id"                       });       models.Document.hasMany         (models.Document_AircraftConfig   ,{as: 'documentsAircraftConfigs' , foreignKey : "document_id"        , targetKey  : 'id'                       });       models.Document.hasMany         (models.Attachment                ,{as: 'attachments'              , foreignKey : "document_id"        , targetKey  : 'id'                       });       models.Document.hasMany         (models.DocumentNote              ,{as: 'notes'                    , foreignKey : "document_id"        , targetKey  : 'id'                       });   

当我询问sequelize.findAll时，我想要有，例如，关于我的主表的15个项目（此处为文档）

当我请求sequelize.find时，我希望在我的主表（此处为文档）上有15个元素，无论选择和连接检索的行数如何。

因此“select”的“limit”子句是不合适的。

我在 PostgreSql 中找到了一种方法：

select *
    
from (
SELECT  dense_rank() over (order by "Document"."id") as "idx","Document"."type_id", "Document"."reference", "Document"."title", "Document"."status_id", "Document"."ata_code", "Document"."ataAH_code", "Document"."issue", "Document"."lead_code", "Document"."documentLinked_id", "Document"."batch_id", "Document"."project_id", "Document"."creationDateAH", "Document"."statusAH", "Document"."budget_id", "Document"."toBeAnalysed", "Document"."createdBy_code", "Document"."updatedBy_code", "Document"."id", "Document"."deleted", "Document"."createdAt", "Document"."updatedAt"
    ,"criterias"."code" AS "criterias.code", "criterias"."title" AS "criterias.title", "criterias"."createdBy_code" AS "criterias.createdBy_code", "criterias"."updatedBy_code" AS "criterias.updatedBy_code", "criterias"."id" AS "criterias.id",  
[...] 
    FROM "Documents" AS "Document" 
    LEFT OUTER JOIN ( "Documents_Criterias" AS "criterias->Document_Criteria" INNER JOIN "Criterias" AS "criterias" ON "criterias"."id" = "criterias->Document_Criteria"."criteria_id") ON "Document"."id" = "criterias->Document_Criteria"."document_id" 
    LEFT OUTER JOIN "Documents_AircraftConfigs" AS "documentsAircraftConfigs" ON "Document"."id" = "documentsAircraftConfigs"."document_id" 
    LEFT OUTER JOIN "Document_Notes" AS "notes" ON "Document"."id" = "notes"."document_id" 
    LEFT OUTER JOIN "Document_Status" AS "status" ON "Document"."status_id" = "status"."id" 
    LEFT OUTER JOIN "Document_Types" AS "type" ON "Document"."type_id" = "type"."id" 
    LEFT OUTER JOIN "Projects" AS "project" ON "Document"."project_id" = "project"."id" 
    LEFT OUTER JOIN "Budgets" AS "budget" ON "Document"."budget_id" = "budget"."id" 
    LEFT OUTER JOIN "Documents" AS "documentLinked" ON "Document"."documentLinked_id" = "documentLinked"."id" 
    LEFT OUTER JOIN "Document_Batches" AS "batch" ON "Document"."batch_id" = "batch"."id" 
    
    WHERE "status"."code" < '25' 
    AND "Document"."deleted" = false 
) as "docs"
    where "idx" <= 15
     

但是如何让 Sequelize 给我这个 sql 命令。

我认为将此功能整合到sequelize中是明智的，例如通过将参数设置为findAll，与“限制”不同，这将使sequelize理解我们只希望在主表上进行限制。并且sequelize将生成一个与当前生成的不同的 sql 顺序。

示例：limitOnMainTable：15

但也许它已经存在了，因为这个功能在分页时非常有用。

您有解决方案可以提供给我吗？

Answer 1

您可以尝试

subQuery: true

，它将限制应用于内部查询（主表）。

例如，假设

Document

为主表，

DocumentType

为子表。以下代码

const result = await Document.findAll({
  include: DocumentType,
  limit: 2,
  subQuery: true
});

将创建一个像这样的查询。

SELECT 
  "Documents".*, 
  "DocumentTypes"."id" AS "DocumentTypes.id", 
  ... 
FROM 
  (
    SELECT 
      "Documents"."id", 
      ...
    FROM 
      "Documents" AS "Documents" 
    WHERE 
      ("Documents"."deletedAt" IS NULL) 
    LIMIT 
      2
  ) AS "Documents" 
  LEFT OUTER JOIN "DocumentTypes" AS "DocumentTypes" ON "Documents"."id" = "DocumentTypes"."DocumentId" 
  AND (
    "DocumentTypes"."deletedAt" IS NULL
  );

limit

应用于主表并获取所有关联的子表的记录。